"problem of inequalities"
Please post only one question at a time.
I will help you solve the first problem.
$\displaystyle \sqrt{3x+1}<x+1$
Square both sides:
$\displaystyle 3x+1<(x+1)^2 =\Rightarrow 3x+1<x^2+2x+1$
$\displaystyle 0<x^2-x$
$\displaystyle 0<x(x-1)$
$\displaystyle 0>x, 1<x$
Only one solution will work in this case
$\displaystyle 1 < x \Rightarrow x > 1$
EDIT: DivideBy0 pointed out a spot that wouldn't work, which I suspected... grr
$\displaystyle 3x+1 \ge 0$
$\displaystyle x \ge \frac{-1}{3}$
I think you made a little typo there which led to a mistake.
$\displaystyle x<0 $ or $\displaystyle x > 1$
But, to be defined,
$\displaystyle 3x+1\geq 0$
$\displaystyle 3x\geq -1$
$\displaystyle x \geq -\frac{1}{3}$
So the solution set is $\displaystyle \{x:-\frac{1}{3}\leq x < 0 \cup x>1\}$
$\displaystyle \frac{x-2}{x+2} > \frac{x^2-4}{x^2+4}$
$\displaystyle \left(\frac{x-2}{x+2}\right)\left(\frac{x-2}{x-2}\right)>\frac{x^2-4}{x^2+4}$
$\displaystyle \frac{(x-2)^2}{x^2-4}>\frac{x^2-4}{x^2+4}$
$\displaystyle \left(\frac{(x-2)^2}{x^2-4}\right)\left(\frac{x^2+4}{x^2+4}\right)>\frac{x^ 2-4}{x^2+4}\left(\frac{x^2-4}{x^2-4}\right)$
$\displaystyle \frac{(x^2+4)(x-2)^2}{x^4-16}>\frac{(x^2-4)^2}{x^4-16}$
Case 1
If $\displaystyle x^4-16>0\Rightarrow x > 2$ or $\displaystyle x<-2$,
$\displaystyle (x^2+4)(x-2)^2>(x^2-4)^2$
$\displaystyle x^4-4x^3+8x^2-16x+16>x^4-8x^2+16$
$\displaystyle 4x^3-16x^2+16x<0$
$\displaystyle 4x(x^2-4x+4)<0$
$\displaystyle 4x(x-2)^2<0$
Which is true so long as $\displaystyle x<0$.
Case 2
If $\displaystyle x^4-16<0\Rightarrow -2 < x< 2$, after some simplification we arrive at
$\displaystyle 4x(x-2)^2>0$
Which is true so long as $\displaystyle x>0$.
So the two regions where the inequality is true is when
$\displaystyle -2<x<2 $ AND $\displaystyle x>0$ $\displaystyle \Rightarrow 0<x<2$.
Or
$\displaystyle x<-2 \cap x>2$ AND $\displaystyle x<0$ $\displaystyle \Rightarrow x<-2$.
So the possible values of x are $\displaystyle \{x:0<x<2 \cup x<-2\}$